[time-nuts] low power divide by 5

[d.schuecker at avm.de]

Hi,

there are three JK-FF, with Q1 as MSB, Q3 as LSB. J1,K1 ist input to Q1, 
etc. .
There are 8^6 possibilties (6 inputs to the Qx or QxNOT or to HIGH or to 
LOW) of which
2069 generate a cycle length of 5.

The following wiring will generate the cycle 1 3 5 2 4 :

J1=Q2
K1=Q1
J2=Q3
K2=Q2
J3=Q2NOT
K3=Q1

Q3 is output with a 3/5 duty cycle.
The unused state 0 will migrate to 1,6->0,7->0.
So the machine does not get stuck.

Cheers
Detlef